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Low iron concentration (as low as 0.02nM) in the open ocean limits phytoplankton growth. Preconcentration is required to determine such low concentrations. Tracefrom a large volume of seawater is concentrated onto achelating resin column. The column is then rinsed with 10mLof 1.5M high-purity water and eluted withofhigh-purityHNO3.

(a) For each sample, seawater is passed through the column for 17hours at 10mL/min. How much is the concentration of Fe3+in the 10mLof NHO3eluate increased by this preconcentration procedure?

(b) What is the concentration of Fe3+in the seawater when 57 nm Fe3+is found in the nitric acid eluate?

(c) Reagent-grade concentrated nitric acid is 15.7 M and contains ≤0.2ppm iron. What would be the apparent concentration of Fe (nM) in a seawater blank if reagent-grade acid were used to prepare the 1.5M HNO3eluent?

Short Answer

Expert verified
  • concentration of Fe3+in the 10mL of HNO3eluate increased = 1020 times
  • concentration of Fe3+in the seawater when 57 nm Fe3+is found in the nitric acid eluate = 0.056nM
  • Apparent concentration of Fe (nM) in a seawater blank if reagent-grade acid were used to prepare the 1.5 MHNO3 eluent = 3.51 nM

Step by step solution

01

a) Calculate the sample volume

V=10mL/min×17h×60min/h/1000mL/L=10.2L=10200mL

Then calculate increase in Fe3concentration

concentration(increased)=totalsamplevolumeelutedsamplevolumeconcentration(increased)=10200mL10mL=1020times

02

b) Now calculate the concentration of Fe3+ in the seawater when 57 nm Fe3+ is found in the nitric acid eluate

c(Fe3+)=c1/concentration(increased)c(Fe3+)=57nM/1020c(Fe3+)=0.056nM

03

c) Then calculate apparent concentration of Fe (nM) in a seawater blank if reagent-grade acid were used to prepare 1.5M HNO3 the eluent

≤200ppb/1020=0.196ppb=0.196×10-9g/mL=0.196×10-6g/L0.196×10-6g/L/55.85g/mol=0.00351×10-6M=3.51nM

Note that 1ppm=1000ppb

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